Question: Evaluate the double integral. $ \int_0^2 \int_{-2x}^{x^2} x - x^2 \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-43}{9}$ (Choice B) B $\dfrac{-9}{2}$ (Choice C) C $\dfrac{-37}{9}$ (Choice D) D $\dfrac{-76}{15}$
Explanation: First, we evaluate the inner integral. We can substitute in the $-2x$ and $x^2$ at the end as if they were numerical bounds. $\begin{aligned} & \int_0^2 \int_{-2x}^{x^2} x - x^2 \, dy \, dx \\ \\ &= \int_0^2 \left[ xy - x^2y \right]_{-2x}^{x^2} \, dx \\ \\ &= \int_0^2 \left(x^3 - x^4 \right) - \left( -2x^2 + 2x^3 \right) \, dx \\ \\ &= \int_0^2 -x^4 - x^3 + 2x^2 \, dx \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^2 -x^4 - x^3 + 2x^2 \, dx &= \left[ \dfrac{-x^5}{5} - \dfrac{x^4}{4} + \dfrac{2x^3}{3} \right]_0^2 \\ \\ &= \dfrac{-32}{5} - \dfrac{16}{4} + \dfrac{16}{3} \\ \\ &= \dfrac{-384 - 240 + 320}{60} \\ \\ &= \dfrac{-304}{60} \\ \\ &= \dfrac{-76}{15} \end{aligned}$ The answer: $ \int_0^2 \int_{-2x}^{x^2} x - x^2 \, dy \, dx = \dfrac{-76}{15}$